我在这里看了算法顺时针旋转,但我不能反过来做。所以基本上,对于顺时针旋转,你需要将转置乘以旋转矩阵,但是如何以相反的方式做同样的事情呢?
这是我的代码:
public class rotation2 {
public static int [][] multiplyMatrix(int [][] m1) {
int [][] m2 = {{0,0,0,1},
{0,0,1,0},
{0,1,0,0},
{1,0,0,0}};
int[][] result = new int[4][4];
// multiply
for (int i=0; i<4; i++)
for (int j=0; j<4; j++)
for (int k=0; k<4; k++)
result[i][j] += m1[i][k] * m2[k][j];
return result;
}
public static int [][] multiplyMatrix2(int [][] m2) {
int [][] m1 = {{0,0,0,1},
{0,0,1,0},
{0,-1,0,0},
{-1,0,0,0}};
int[][] result = new int[4][4];
// multiply
for (int i=0; i<4; i++)
for (int j=0; j<4; j++)
for (int k=0; k<4; k++)
result[i][j] += m1[i][k] * m2[k][j];
return result;
}
public static void printArray(int [][] array) {
for(int row = 0; row < array.length; row++) {
for(int col = 0; col < array[row].length; col++) {
if (array[row][col] > 0) {
System.out.printf("1");
} else {
System.out.printf("0");
}
}
System.out.printf("\n");
}
}
public static int [][] transpose(int [][] m1) {
int m = 4;
int n = 4;
int c = 0;
int d = 0;
int[][] transpose = new int [n][m];
for ( c = 0 ; c < m ; c++ ) {
for ( d = 0 ; d < n ; d++ ) {
transpose[d][c] = m1[c][d];
}
}
return transpose;
}
public static void main(String[] args) {
int [][] m1 = {{1,0,0,0},
{1,0,0,0},
{1,1,0,0},
{0,0,0,0}};
int [][] transpose = transpose(m1);
printArray(transpose);
transpose = multiplyMatrix(transpose);
printArray(transpose);
int [][] transpose2 = transpose(m1);
printArray(transpose2);
transpose2 = multiplyMatrix(transpose2);
printArray(transpose2);
}
}
你没有转换计数器时钟旋转,对吗?
答案 0 :(得分:0)
你要做的是(1)转置矩阵,(2)分别反转行(顺时针)或列(逆时针)。
您可以使用双循环来设置单个单元格的新值,同时执行这两个步骤。在代码中,这可能如下所示:
public static int[][] rotate(int[][] m, boolean left) {
int rows = m.length, cols = m[0].length;
int[][] m2 = new int[cols][rows]; // swap rows and cols
for (int r = 0; r < rows; r++)
for (int c = 0; c < cols; c++)
if (left) // rotate left
m2[c][r] = m[r][cols - c - 1];
else // rotate right
m2[c][r] = m[rows - r - 1][c];
return m2;
}
有关更多信息和替代方法,请查看this related question的答案。